If `-1 le x le 0` then `cos^(-1)(2x^(2)-1)` equals

Options

A. `2 cos^(-1)x`
B. `pi-2 cos^(-1)x`
C. `2pi-2 cos^(-1)x`
D. `-2 cos^(-1)x`

Answer

To solve the problem, we need to find the value of \( \cos^{-1}(2x^2 - 1) \) given that \( -1 \leq x \leq 0 \).

### Step-by-Step Solution:

1. **Understanding the Range of \( x \)**:
We are given that \( x \) lies between -1 and 0, i.e., \( -1 \leq x \leq 0 \).

2. **Using the Trigonometric Identity**:
We can use the identity:
\[ \cos^{-1}(2x^2 - 1) = 2 \cos^{-1}(x) \]
This identity holds true for \( x \) in the range of \( [0, 1] \). However, since \( x \) is in the range of \( [-1, 0] \), we need to adjust our approach.

3. **Finding \( \cos^{-1}(2x^2 - 1) \)**:
Since \( x \) is negative, we can express \( x \) as \( -y \) where \( 0 \leq y \leq 1 \). Thus, we have:
\[ 2x^2 - 1 = 2(-y)^2 - 1 = 2y^2 - 1 \]

4. **Applying the Identity**:
Now, we can apply the identity:
\[ \cos^{-1}(2y^2 - 1) = 2 \cos^{-1}(y) \]
Since \( y = -x \), we can write:
\[ \cos^{-1}(2x^2 - 1) = 2 \cos^{-1}(-x) \]

5. **Finding \( \cos^{-1}(-x) \)**:
For \( -1 \leq x \leq 0 \), \( -x \) lies between 0 and 1. Therefore, \( \cos^{-1}(-x) \) is simply:
\[ \cos^{-1}(-x) = \pi - \cos^{-1}(x) \]

6. **Final Expression**:
Substituting back, we have:
\[ \cos^{-1}(2x^2 - 1) = 2(\pi - \cos^{-1}(x)) = 2\pi - 2\cos^{-1}(x) \]

### Conclusion:
Thus, the final answer is:
\[ \cos^{-1}(2x^2 - 1) = 2\pi - 2\cos^{-1}(x) \]